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sql联合查询语句

select C.stuid,C.Maxscore as score,B.coursename,B.status from (select A.stuid,max(B.score) as Maxscore from A,B where A.stuid=B.stuid group by A.stuid) C,B where C.stuid=B.stuid and C.Maxscore =B.score 若还有问题,请Hi我~

不只是联合啊,还要行列转换 declare @sql varchar(1000) set @sql='select 姓名' select @sql=@sql+',max(case 课程 when ''' + 课程 + ''' then 分数 else 0 end) ['+课程+']' from (select distinct 课程 from (select a.nname 姓名,b.fname 课程,c.score

Select id, Sum(Num) As 总数 From ((Select id, num1 As NumFrom A表) Union (Select id, num2 As Num From B表)) Group By id, Num; 汇总是在子查询基础上建立的,联合用个Union就好了,汇总一定要给他Group By,做分组

select t1.id,t1.name,t1.数量,t2.ImgUrl from (select a.id,a.name,count(1) as 数量,max(paixu) as paixu_id from user a,images bwhere a.id=b.tidgroup by a.id,a.name) t1 left join images t2 on t1.id=t2.tid and t1.paixu_id=t2.paixu

select a.id,a.name,b.name,c.name,a.table2_id,a.table3_id,c.table2_id from table1 a,table2 b,table3 c where a.table2_id=b.id and a.table3_id=c.id and b.id=c.table2_id order by a.id; 以上语句在Oracle11g r2上测试通过,同样可以用于其他数据库,如有疑问请留言

都有只取一个 用这个select a1,b1,c1 from table1unionselect a1,b1,c1 from table2;如果都有,不去重 用这个select a1,b1,c1 from table1union allselect a1,b1,c1 from table2;

select*from(selecta.姓名编号,a.姓名,b.考试科目as科目,b.成绩,row_number()over(partitionbya.姓名orderbyb.成绩desc)asidfrom表一a,表二bwherea.姓名=b.姓名)whereid=1;说明:1、上面的sql是按照成绩降序排列后取出每个姓名的第一条记录,如果想换成升序把【orderbyb.成绩desc】改成【orderbyb.成绩】即可.2、如果想取第二条记录把【whereid=1;】改成【whereid=2;】即可.3、如果想去前n条把【whereid=1;】改成【whereid<=n;】即可.---以上,希望对你有所帮助.

select a.no,a.name,b.subid,b.subname,c.score from a,b,c where a.no = c.no and b.subid = c.subid ;

应该是GROUP BY的问题是按照[Check].CheckDate, [Check].CheckID, [Check].Customer_ID来分组,也就是说[Check].CheckDate, [Check].CheckID, [Check].Customer_ID这三个条件联合分组,你看看你的查询结果这三个条件组合在一起是不是唯一的看看GROUP BY的相关资料吧,我建议group by 条件少点

select x.userid,x.distyname,y.dealname from (select userid,name as distyname from b left join a on b.cust_id=a.id where a.cust_type=1) x left join (select userid,name as dealname from b left join a on b.cust_id=a.id where a.cust_type=2) y on x.userid=y.userid

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